# First law of therodynamics appiled across a normal shock

This is a snippet from the book Compressible Flow by Anderson. Here, he is trying to evaluate the change in entropy across the shock using the relation,

$$s_2 - s_1 = c_p \ ln \frac{T_2}{T_1} - R \ ln \frac{P_2}{P_1}$$.

This equation has been obtained using the first law,

$$\delta q = de + Pdv$$, where $$e$$ is just the internal energy of the fluid. This is how many of the books on Compressible flow (Shapiro, Liepman etc) derive it. My question is, how can we ignore the contribution of the kinetic energy to the first law, i.e, why isn't first law,

$$\delta q = d (e + \frac{u^2}{2}) + Pdv$$.

by

• In the more general form of the first law, kinetic energy and gravitational potential energy are included. – Chet Miller May 18 at 16:22

There are two forms of kinetic energy of a system: microscopic and macroscopic.

Microscopic kinetic energy is part of the internal energy of a system and is reflected by the systems temperature.

Macroscopic kinetic energy is associated with the velocity of the system as a whole. It’s the system’s external kinetic energy (rotational and translational) with respect to an external frame of reference.

Are we ignoring the contribution of the macroscopic kinetic energy in the first law?

Yes

The macroscopic kinetic energy of the system has no impact on the internal energy of the system.

Say you have a perfectly insulated rigid box of an ideal gas. The internal energy of an ideal gas depends only on its temperature. You measure the temperature of the gas as being $$T$$ in your laboratory. Now take the box with you in your car which is moving at a velocity $$v$$ with respect to the road. The box of gas also has a velocity $$v$$ with respect to the road and therefore has a macroscopic kinetic energy of $$\frac{mv^2}{2}$$ with respect to the road (a frame of reference external to the gas). Measure the temperature of the gas again. Would you expect the temperature of the gas to increase? No. Therefore the macroscopic kinetic energy of the gas has no impact on its internal energy.

As an explosion is taking place and the explosion front is moving outward, it accelerates the fluid it comes in contact with in the direction of its propagation and thus increases the surrounding fluid's KE

If the fluid is accelerated by a force, then there is more involved than simply increasing the macroscopic kinetic energy of the fluid. The change in macroscopic kinetic energy, $$\frac{mv^2}{2}$$, is still not part of the change in internal energy. However, if the fluid is compressible, the shock wave will do work on the fluid by compressing it, just as if we were compressing a gas inside a cylinder with a piston. In that case, the explosion DOES increase the internal energy of the fluid but that increase is already included in the $$Pdv$$ term of your equation.

The $$Pdv$$ work done on the fluid by the explosion increases the internal kinetic energy of the fluid and is reflected by an increase in temperature so that $$T_{2}>T_{1}$$ in your entropy equation. Since the explosion also increases the velocity of the fluid as a whole, it increases the external kinetic energy of the fluid. But this does not cause a temperature increase and has no influence on internal energy or the change in entropy.

The increase in pressure of the fluid increases its density and therefore increases the “orderliness” of the molecules. This results in a decrease in entropy and is consistent with the fact that, all other things being equal, for a given substance, the entropy of a solid is less than of a liquid which, in turn, is less than that of a gas. In your change in entropy equation, $$P_{2}>P_{1}$$ in the second term, and therefore reduces the entropy.

Hope this helps.

• Comments are not for extended discussion; this conversation has been moved to chat. – Chris May 19 at 21:59